The accepted (true) density of aluminum is\u00c3\u201a\u00c2 2.70\u00c3\u201a\u00c2 g\/cm3. If a student, after performing an experiment, found the value to be\u00c3\u201a\u00c2 2.45\u00c3\u201a\u00c2 g\/cm3, calculate the percent error for this value. The accepted (true) density of aluminum is\u00c3\u201a\u00c2 2.70\u00c3\u201a\u00c2 g\/cm3. If a student, after performing an experiment, found the value to be\u00c3\u201a\u00c2 2.45\u00c3\u201a\u00c2 g\/cm3, calculate the percent error for this value. Group of answer choices \u00c3\u201a\u00c2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the percent error, we use the following formula:<\/p>\n\n\n\n [ In this case, the true (accepted) density of aluminum is given as 2.70 g\/cm\u00b3, and the experimental density found by the student is 2.45 g\/cm\u00b3.<\/p>\n\n\n\n [ [ [ [ So, the percent error is 9.26%<\/strong>.<\/p>\n\n\n\n Percent error is a way of determining how close an experimental value is to the true or accepted value. In this case, the student found a density of 2.45 g\/cm\u00b3 for aluminum, which is slightly lower than the accepted value of 2.70 g\/cm\u00b3. The negative sign in the difference indicates that the experimental value is lower than the true value, but since percent error is always expressed as a positive value, we take the absolute value of the difference.<\/p>\n\n\n\n The formula used for percent error shows the relative size of the difference between the experimental and true values as a percentage of the true value. In this case, the difference is small relative to the true value (approximately 9.26%), indicating that the student’s experimental result is reasonably close to the true value, though not exact.<\/p>\n\n\n\n Percent error is important in scientific experiments because it helps to assess the precision of the measurements and whether any systematic errors may have affected the results. A larger percent error would indicate that the experimental value deviates significantly from the true value, while a smaller percent error indicates more accurate results.<\/p>\n","protected":false},"excerpt":{"rendered":" The accepted (true) density of aluminum is\u00c3\u201a\u00c2 2.70\u00c3\u201a\u00c2 g\/cm3. If a student, after performing an experiment, found the value to be\u00c3\u201a\u00c2 2.45\u00c3\u201a\u00c2 g\/cm3, calculate the percent error for this value. The accepted (true) density of aluminum is\u00c3\u201a\u00c2 2.70\u00c3\u201a\u00c2 g\/cm3. If a student, after performing an experiment, found the value to be\u00c3\u201a\u00c2 2.45\u00c3\u201a\u00c2 g\/cm3, calculate the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-182597","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182597","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=182597"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182597\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182597"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182597"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182597"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{Percent Error} = \\left| \\frac{\\text{Experimental Value} – \\text{True Value}}{\\text{True Value}} \\right| \\times 100
]<\/p>\n\n\n\nStep 1: Find the difference between the experimental and true values.<\/h3>\n\n\n\n
\\text{Difference} = 2.45 \\, \\text{g\/cm}^3 – 2.70 \\, \\text{g\/cm}^3 = -0.25 \\, \\text{g\/cm}^3
]<\/p>\n\n\n\nStep 2: Take the absolute value of the difference.<\/h3>\n\n\n\n
\\left| -0.25 \\, \\text{g\/cm}^3 \\right| = 0.25 \\, \\text{g\/cm}^3
]<\/p>\n\n\n\nStep 3: Divide the difference by the true value.<\/h3>\n\n\n\n
\\frac{0.25 \\, \\text{g\/cm}^3}{2.70 \\, \\text{g\/cm}^3} = 0.0926
]<\/p>\n\n\n\nStep 4: Multiply by 100 to get the percentage.<\/h3>\n\n\n\n
0.0926 \\times 100 = 9.26\\%
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n